In the tetrahedron $SABC $ at the height $SH$ the following point $O$ is chosen, such that: $$\angle AOS + \alpha = \angle BOS + \beta = \angle COS + \gamma = 180^o, $$where $\alpha, \beta, \gamma$ are dihedral angles at the edges $BC, AC, AB $, respectively, at this point $H$ lies inside the base $ABC$. Let ${{A} _ {1}}, \, {{B} _ {1}}, \, {{C} _ {1}} $be the points of intersection of lines and planes: ${{A} _ {1}} = AO \cap SBC $, ${{B} _ {1}} = BO \cap SAC $, ${{C} _ {1}} = CO \cap SBA$ . Prove that if the planes $ABC $ and ${{A} _ {1}} {{B} _ {1}} {{C} _ {1}} $ are parallel, then $SA = SB = SC $. (Alexey Klurman)